{"id":26024,"date":"2025-10-03T19:19:27","date_gmt":"2025-10-03T13:49:27","guid":{"rendered":"https:\/\/onlinefreenotes.com\/?p=26024"},"modified":"2025-12-12T07:11:43","modified_gmt":"2025-12-12T07:11:43","slug":"spectrum-icse-class-10-physics-notes","status":"publish","type":"post","link":"https:\/\/mockupbw.site\/2025\/onlinefreenotes\/spectrum-icse-class-10-physics-notes\/","title":{"rendered":"Spectrum: ICSE Class 10 Physics solutions, notes"},"content":{"rendered":"\n<p>Get summaries, questions, answers, solutions, notes, extras, Concise\/Selina workbook solutions, PDF and guide of chapter 6 Spectrum: ICSE Class 10 Physics which is part of the syllabus of students studying under the <a href=\"http:\/\/Council for the Indian School Certificate Examinations\">Council for the Indian School Certificate Examinations<\/a> board. These solutions, however, should only be treated as references and can be modified\/changed.&nbsp;<\/p>\n\n\n  <style>\r\n    .notice {\r\n      background: yellow;       \/* simple yellow background *\/\r\n      text-align: center;       \/* centre alignment *\/\r\n      padding: 12px 16px;\r\n      margin: 20px auto;\r\n      width: fit-content;       \/* shrink to text and centre via auto margins *\/\r\n      font-family: Arial, sans-serif;\r\n    }\r\n  <\/style>\r\n  <div class=\"notice\">\r\n    If you notice any errors in the notes, please mention them in the comments\r\n  <\/div>\r\n<nav id=\"toc\" class=\"toc-box\"><\/nav>\r\n<style>\r\n.toc-box{\r\n  border:1px solid #e5e7eb;\r\n  border-radius:8px;\r\n  background:#fff;\r\n  margin:20px 0;\r\n  font-family:Arial, Helvetica, sans-serif\r\n}\r\n.toc-header{\r\n  padding:10px 14px;\r\n  font-size:16px;\r\n  font-weight:600;\r\n  border-bottom:1px solid #eef2f7;\r\n  background:#f8fafc\r\n}\r\n.toc-content{\r\n  padding:12px 18px\r\n}\r\n\r\n\/* Base list *\/\r\n.toc-content ul{\r\n  margin:0 25px;\r\n  padding-left:0;\r\n  list-style:none\r\n}\r\n\r\n\/* Level-based bullets *\/\r\n.toc-content li{\r\n  position:relative;\r\n  margin:6px 0;\r\n  margin-left:6px;\r\n  line-height:1.5;\r\n\tlist-style:disc;\r\n}\r\n\r\n\/* H2 bullet \u25cf *\/\r\n.toc-content li.level-2{\r\n  list-style:disc;\r\n\t\r\n}\r\n\r\n\/* H3 bullet \u25cb *\/\r\n.toc-content li.level-3{\r\n  margin-left:26px;\r\n\tlist-style:disc;\r\n}\r\n\r\n\r\n\/* H4+ bullet \u2013 *\/\r\n.toc-content li.level-4{\r\n  margin-left:46px;\r\n\tlist-style:disc;\r\n}\r\n.toc-content li.level-5,\r\n.toc-content li.level-6{\r\n  margin-left:66px;\r\n\tlist-style:disc;\r\n}\r\n\r\n.toc-content a{\r\n  text-decoration:none;\r\n  color:#000\r\n}\r\n.toc-content a:hover{\r\n  text-decoration:underline\r\n}\r\n\r\nhtml{scroll-behavior:smooth}\r\nh1[id],h2[id],h3[id],h4[id],h5[id],h6[id]{\r\n  scroll-margin-top:110px\r\n}\r\n<\/style>\r\n\r\n<script>\r\ndocument.addEventListener('DOMContentLoaded', function () {\r\n\r\n  const toc = document.getElementById('toc');\r\n  if (!toc) return;\r\n\r\n  \/* MAIN CONTENT ONLY *\/\r\n  const content = document.querySelector('#pdf-content');\r\n\r\n  \/* EXCLUDE AREAS *\/\r\n  const excludeSelectors = `\r\n    .author, .byline, .entry-meta, .post-meta,\r\n    #comments, .comments-area, .comment-respond,\r\n    .comment-form, .comment-list,\r\n    .login, .login-required,\r\n    .sidebar, aside, footer, nav,\r\n    .widget, .widgets\r\n  `;\r\n\r\n  \/* TEXT TO IGNORE *\/\r\n  const ignoreText = [\r\n    'leave a comment',\r\n    'cancel reply',\r\n    'login required',\r\n    'get notes',\r\n    'ron\\'e dutta',\r\n    'comments'\r\n  ];\r\n\r\n  \r\nconst headings = [...content.querySelectorAll('h1,h2,h3,h4,h5,h6')]\r\n  .filter(h => !excludeSelectors || !h.closest(excludeSelectors))\r\n  .filter(h => {\r\n    const txt = h.textContent.trim().toLowerCase();\r\n    return txt.length > 0 && !ignoreText.some(t => txt.includes(t));\r\n  });\r\n\r\n\/\/alert(content);\r\n  if (!headings.length) {\r\n    toc.style.display = 'none';\r\n    return;\r\n  }\r\n\r\n  \/* UNIQUE IDs *\/\r\n  const used = {};\r\n  const slug = t => t.toLowerCase().trim()\r\n    .replace(\/[^a-z0-9\\s-]\/g, '')\r\n    .replace(\/\\s+\/g, '-');\r\n\r\n  headings.forEach(h => {\r\n    if (!h.id) {\r\n      let base = slug(h.textContent) || 'section';\r\n      used[base] = (used[base] || 0) + 1;\r\n      h.id = used[base] > 1 ? base + '-' + used[base] : base;\r\n    }\r\n  });\r\n\r\n  \/* BUILD TOC *\/\r\n  const ul = document.createElement('ul');\r\n\r\n  headings.forEach(h => {\r\n    const level = parseInt(h.tagName.substring(1));\r\n    if (level < 2) return; \/\/ skip H1 like your reference site\r\n\r\n    const li = document.createElement('li');\r\n    li.className = 'level-' + level;\r\n\r\n    const a = document.createElement('a');\r\n    a.href = '#' + h.id;\r\n    a.textContent = h.textContent.trim();\r\n\r\n    li.appendChild(a);\r\n    ul.appendChild(li);\r\n  });\r\n\r\n  toc.innerHTML = `\r\n    <div class=\"toc-header\">Table of Contents<\/div>\r\n    <div class=\"toc-content\"><\/div>\r\n  `;\r\n  toc.querySelector('.toc-content').appendChild(ul);\r\n\r\n});\r\n<\/script>\r\n\n\n\n\n<h3 class=\"wp-block-heading\" id=\"Summary\"><strong>Summary<\/strong><\/h3>\n\n\n\n<p>When light passes through a special glass shape called a prism, it bends. This bending is called deviation. How much the light bends depends on a few things: the angle at which light hits the prism, the shape of the prism, and the material the prism is made from. It also changes with the color of the light. You see, what we call white light is actually a mix of many different colors, like those you see in a rainbow.<\/p>\n\n\n\n<p>As white light goes into a prism, each color inside it bends by a different amount. This happens because each color travels at a slightly different speed when it&#8217;s inside the prism. Violet light, which has a shorter wavelength (you can think of wavelength as the length of a light wave&#8217;s &#8220;step&#8221;), travels the slowest and so it bends the most. Red light, which has a longer &#8220;step,&#8221; travels the fastest and bends the least. This act of splitting white light into its different colors \u2013 violet, indigo, blue, green, yellow, orange, and red, often remembered as VIBGYOR \u2013 is called dispersion. The beautiful band of colors that appears is known as a spectrum. A famous scientist, Sir Isaac Newton, was one of the first to show this. The main separation of colors happens when the light first enters the prism.<\/p>\n\n\n\n<p>The light that our eyes can see is just a tiny piece of a much larger family of waves called the electromagnetic spectrum. This big family also includes waves that are invisible to us. Waves that have &#8220;steps&#8221; longer than red light include infrared waves (which we feel as heat), microwaves (the kind used in microwave ovens), and radio waves (which carry TV and radio signals). On the other side, waves with &#8220;steps&#8221; shorter than violet light include ultraviolet waves (which come from the sun and can cause sunburns), X-rays (which doctors use to see inside our bodies, like our bones), and gamma rays (which are very energetic). All these different waves travel at the same incredibly fast speed through empty space or air. They don&#8217;t need anything to travel through, and they can all bounce off things (that&#8217;s reflection) and bend when they pass from one material to another (that&#8217;s refraction).<\/p>\n\n\n\n<p>Gamma rays have the shortest &#8220;steps&#8221; and are very powerful. X-rays can go through the soft parts of our body but are stopped by harder parts like bones, which is why they are useful for medical images. Ultraviolet (UV) light can be harmful if we get too much of it, but a little bit helps our bodies make vitamin D. We can find UV light because it causes changes in some special chemicals. Infrared (IR) waves are what we feel as heat. They are used in things like TV remote controls and for seeing in the dark. Radio waves have the longest &#8220;steps&#8221; of all.<\/p>\n\n\n\n<p>Sunlight also does something interesting called scattering when it bumps into tiny dust particles and air molecules in our planet&#8217;s atmosphere. Light with shorter &#8220;steps,&#8221; like blue and violet, scatters much more easily than light with longer &#8220;steps,&#8221; like red. This is the reason the sky usually looks blue. The blue part of sunlight scatters all around in the air and comes to our eyes from every direction. When the sun is rising or setting, its light has to travel a much longer way through the atmosphere to reach us. By the time it gets to our eyes, most of the blue light has been scattered away, so the sun and the sky around it look red. Clouds appear white because the water droplets in them are much bigger than air molecules, and they scatter all the colours of sunlight almost equally. Danger signals are often red because red light scatters the least, so it can be seen clearly from a long distance, even in hazy weather.<\/p>\n\n\n\n\n\n<h3 class=\"wp-block-heading\" id=\"Very_short_answers\"><strong>Workbook solutions<\/strong> (Concise\/Selina)<\/h3>\n\n\n\n<h4 class=\"wp-block-heading\" id=\"Very_Short_Answer_Type_Questions\"><strong>EXERCISE (A)<\/strong><\/h4>\n\n\n\n<h5 class=\"wp-block-heading\" id=\"Very_Short_Answer_Type_Questions\"><strong>MCQ<\/strong><\/h5>\n\n\n\n<p><strong>1. When a white light ray falls on a prism, the ray at its first surface suffers :<\/strong><\/p>\n\n\n\n<p>(a) no refraction<br>(b) only dispersion<br>(c) only deviation<br>(d) both deviation and dispersion.<\/p>\n\n\n\n<p><strong>Answer:<\/strong> (d) both deviation and dispersion.<\/p>\n\n\n\n<p><strong>2. When a ray of white light falls on a prism, which of the following statements are correct ?<\/strong><\/p>\n\n\n\n<p>(a) The dispersion of white light occurs at the first surface of the prism.<br>(b) The deviation of light rays occurs at both the surfaces of the prism.<br>(c) The prism does not produce colours, but it only splits the various colours present in the white light.<br>(d) All of the above.<\/p>\n\n\n\n<p><strong>Answer:<\/strong> (d) All of the above.<\/p>\n\n\n\n<p><strong>3. The correct arrangement of colours in an increasing order of their wavelengths is :<\/strong><\/p>\n\n\n\n<p>(a) Violet &lt; Green &lt; Red<br>(b) Red &lt; Green &lt; Violet<br>(c) Green &lt; Violet &lt; Red<br>(d) Green &lt; Red &lt; Violet<\/p>\n\n\n\n<p><strong>Answer:<\/strong> (a) Violet &lt; Green &lt; Red<\/p>\n\n\n\n<p><strong>4. Out of red, blue and violet, which colour has the greatest speed in vacuum ?<\/strong><\/p>\n\n\n\n<p>(a) red<br>(b) blue<br>(c) violet<br>(d) all have the same speed<\/p>\n\n\n\n<p><strong>Answer:<\/strong> (d) all have the same speed<\/p>\n\n\n\n<p><strong>5. A beam consisting of red, blue and yellow colours is incident normally on the face AB of an isosceles right-angled prism ABC as shown in the figure given below. Critical angle of glass-air interface for yellow colour is 45\u00b0. Out of the four emergent rays P, Q, R and S, which one is for the yellow colour :<\/strong><\/p>\n\n\n\n<img decoding=\"async\" src=\"https:\/\/blogger.googleusercontent.com\/img\/b\/R29vZ2xl\/AVvXsEjCOn1Y1AIH7BZv2Hyqa6fcTW97Xxln7YIWuckavyzc-chVxazwEdZ4FATIUurYUtMCMsjmoxcVv3cNfrfPQ8Tmg_HgXZTVws08D-qxNhgeLhFDqPUvbjADwuCkhQwrlgaaoDjVQt6JxR4F5jkV6hDwBgJUgq2OlI7hN2YlmEqfTeEzC_DLyjtbxXa91NNN\/s1472\/6.1.png\" style=\"max-width: 100%; height: auto; margin-bottom: 20px;\" \/>\n\n\n\n<p>(a) P<br>(b) Q<br>(c) R<br>(d) S<\/p>\n\n\n\n<p><strong>Answer:<\/strong> (b) Q<\/p>\n\n\n\n<p><strong>6. The frequency of violet light is 7-5 x 10\u00b9\u2074 Hz. Its wavelength in nm is:<\/strong><\/p>\n\n\n\n<p>(a) 7500 nm<br>(b) 4000 nm<br>(c) 400 nm<br>(d) 750 nm<\/p>\n\n\n\n<p><strong>Answer:<\/strong> (c) 400 nm<\/p>\n\n\n\n<p><strong>7. The correct order of angle of deviation of indigo, green, yellow and red colours is:<\/strong><\/p>\n\n\n\n<p>(a) \u03b4\u1d62 &gt; \u03b4g &gt; \u03b4y &gt; \u03b4\u1d63<br>(b) \u03b4\u1d62 &lt; \u03b4g &lt; \u03b4y &lt; \u03b4\u1d63<br>(c) \u03b4g &gt; \u03b4\u1d62 &gt; \u03b4y &gt; \u03b4\u1d63<br>(d) \u03b4\u1d63 &gt; \u03b4y &gt; \u03b4g &gt; \u03b4\u1d62<\/p>\n\n\n\n<p><strong>Answer:<\/strong> (a) \u03b4\u1d62 &gt; \u03b4g &gt; \u03b4y &gt; \u03b4\u1d63<\/p>\n\n\n\n<p><strong>8. Assertion (A): When a ray of light is refracted through a rectangular glass slab, there is no dispersion of light.<\/strong><strong><br><\/strong><strong>Reason (R): Dispersion of light is the phenomenon of splitting of white light into its constituent colours.<\/strong><\/p>\n\n\n\n<p>(a) both A and R are true and R is the correct explaination of A<br>(b) both A and R are True and R is not the correct explanation of A<br>(c) assertion is false but reason is true<br>(d) assertion is true but reason is false<\/p>\n\n\n\n<p><strong>Answer:<\/strong> (b) both A and R are True and R is not the correct explanation of A<\/p>\n\n\n\n<p><strong>9. Assertion (A): A beam of white light gives a spectrum on passing through a hollow prism.<\/strong><strong><br><\/strong><strong>Reason (R): The speed of light outside the prism is same as the speed of light inside the prism.<\/strong><\/p>\n\n\n\n<p>(a) both A and R are true and R is the correct explaination of A<br>(b) both A and R are True and R is not the correct explanation of A<br>(c) assertion is false but reason is true<br>(d) assertion is true but reason is false<\/p>\n\n\n\n<p><strong>Answer:<\/strong> (c) assertion is false but reason is true<\/p>\n\n\n\n<h5 class=\"wp-block-heading\" id=\"Very_Short_Answer_Type_Questions\"><strong>Very Short Answer Type Questions<\/strong><\/h5>\n\n\n\n<p><strong>1. How does the speed of light in glass change on increasing the wavelength of light?<\/strong><br><strong><br>Answer<\/strong>: The speed of light in glass increases with an increase in its wavelength.<\/p>\n\n\n\n<p><strong>2. Which colors of white light travels (a) fastest (b) slowest, in glass?<\/strong><br><strong><br>Answer<\/strong>: (a) In glass, red light travels fastest.<br>(b) In glass, violet light travels slowest.<\/p>\n\n\n\n<p><strong>3. Name the subjective property of light related to its wavelength.<\/strong><br><strong><br>Answer<\/strong>: The subjective property of light related to its wavelength is colour.<\/p>\n\n\n\n<p><strong>4. What is the range of wavelength of the spectrum of white light in (i) \u00c5, (ii) nm?<\/strong><br><strong><br>Answer<\/strong>: The range of wavelength of the spectrum of white light is:<br>(i) 4000 \u00c5 to 8000 \u00c5<br>(ii) 400 nm to 800 nm<\/p>\n\n\n\n<p><strong>5. Name four colors of the spectrum of white light which have wavelength longer than blue light.<\/strong><br><strong><br>Answer<\/strong>: Four colors of the spectrum of white light which have wavelengths longer than blue light are green, yellow, orange, and red.<\/p>\n\n\n\n<p><strong>6. How will the dispersion of light through a prism change if the prism is made of diamond instead of glass?<\/strong><br><strong><br>Answer<\/strong>: If the prism is made of diamond instead of glass, there will be a greater angular deviation of all colours, implying a greater dispersion.<\/p>\n\n\n\n<h5 class=\"wp-block-heading\" id=\"Short_Answer_Type_Questions:\"><strong>Short Answer Type Questions<\/strong><\/h5>\n\n\n\n<p><strong>1. Name three factors on which the deviation produced by a prism depends and state how does it depend on the factors stated by you.<\/strong><br><strong><br>Answer<\/strong>: The total angle of deviation \u03b4 produced by a prism depends upon the following three factors:<\/p>\n\n\n\n<p>(1) The angle of incidence (i) at the first surface. The angle of deviation depends on the angle of incidence.<br>(2) The angle of the prism (A). The angle of deviation depends on the angle of the prism.<br>(3) The refractive index of the material of the prism (\u03bc). The angle of deviation depends on the refractive index. Since the refractive index depends on the color (or wavelength \u03bb) of the light used, the angle of deviation also depends on the colour (or wavelength \u03bb) of the incident light. Specifically, the deviation caused by a prism increases with the decrease in the wavelength of light.<\/p>\n\n\n\n<p><strong>2. How does the deviation produced by a triangular prism depend on the colour (or wavelength) of light incident on it?<\/strong><br><strong><br>Answer<\/strong>: The deviation produced by a triangular prism increases with the decrease in the wavelength of light incident on it. In visible light, violet colour (shortest wavelength) is deviated the most and red colour (longest wavelength) is deviated the least.<\/p>\n\n\n\n<p><strong>3. (a) Write the approximate wavelength for (i) blue, and (ii) red light.<br>(b) The wavelength of violet and red light are 4000\u00c5 and 8000\u00c5 respectively. Which of the two has higher frequency?<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong>: (a) (i) The approximate wavelength for blue light is 4800 \u00c5.<br>(ii) The approximate wavelength for red light is 8000 \u00c5.<br>(b) Violet light of wavelength 4000\u00c5 has the higher frequency.<\/p>\n\n\n\n<p><strong>4. Write the seven prominent colours present in white light in the order of increasing wavelength.<\/strong><br><strong><br>Answer<\/strong>: The seven prominent colours present in white light, in the order of increasing wavelength, are: violet, indigo, blue, green, yellow, orange, and red.<\/p>\n\n\n\n<p><strong>5. Name the seven prominent colours of the white light spectrum in order of their increasing frequencies.<\/strong><br><strong><br>Answer<\/strong>: The seven prominent colours of the white light spectrum, in order of their increasing frequencies, are: red, orange, yellow, green, blue, indigo, and violet.<\/p>\n\n\n\n<p><strong>6. The wavelengths for the light of red and blue colours are nearly 7.8 \u00d7 10\u207b\u2077 m and 4.8 \u00d7 10\u207b\u2077 m respectively.<\/strong><\/p>\n\n\n\n<p><strong>(a) Which colour has greater speed in vacuum?<br>(b) Which colour has greater speed in glass?<\/strong><br><strong><br>Answer<\/strong>: (a) In vacuum, both red and blue colours have the same speed.<br>(b) In glass, red light has a greater speed.<\/p>\n\n\n\n<p><strong>7. Define the term dispersion of light.<\/strong><br><strong><br>Answer<\/strong>: Dispersion of light is the phenomenon of splitting of white light by a prism into its constituent colours.<\/p>\n\n\n\n<p><strong>8. What do you understand by the term spectrum?<\/strong><br><strong><br>Answer<\/strong>: Spectrum is the band of colours seen on a screen when white light is passed through a prism.<\/p>\n\n\n\n<p><strong>9. A ray of white light is passed through a glass prism and a spectrum is obtained on a screen.<\/strong><\/p>\n\n\n\n<p><strong>(a) Name the seven colours of the spectrum in order.<br>(b) Do the colours have the same width in the spectrum?<br>(c) Which colour of the spectrum of white light deviates (i) the most, (ii) the least?<\/strong><br><strong><br>Answer<\/strong>: (a) Starting from the side of the base of the prism, the seven colours of the spectrum in order are: Violet, Indigo, Blue, Green, Yellow, Orange, and Red.<\/p>\n\n\n\n<p>(b) No, the colours do not have the same width in the spectrum; different colours have different width on the screen.<\/p>\n\n\n\n<p>(c) (i) Violet colour of the spectrum of white light deviates the most.<br>(ii) Red colour of the spectrum of white light deviates the least.<\/p>\n\n\n\n<p><strong>10. How would the dispersion of light change if the prism is submerged in water instead of air? Will the spread of the constituent colours remain the same?<\/strong><br><strong><br>Answer<\/strong>: If the prism is submerged in water instead of air, the refractive index of the prism material with respect to the surrounding medium (water) decreases. Since the refractive index of water is higher than air, the deviation of light would be less, thereby reducing the dispersion effect and thus the spread of constituent colours would be less pronounced. The spread of constituent colours will not remain the same; it will be less.<\/p>\n\n\n\n<h5 class=\"wp-block-heading\" id=\"Long_Answer_Type_Questions\"><strong>Long Answer Type Questions<\/strong><\/h5>\n\n\n\n<p><strong>1. Explain the cause of dispersion of white light through a prism.<\/strong><br><strong><br>Answer<\/strong>: The cause of dispersion of white light is that light of different wavelengths (colours) travel with different speeds in a medium other than air or vacuum. When white light enters the first surface of a prism, lights of different colours, due to their different speeds in glass, get deviated through different angles towards the base of the prism. This splitting of white light into its constituent colours, known as dispersion, takes place at the first surface of the prism. Violet colour, having the shortest wavelength, travels slowest in glass and is deviated the most, while red colour, having the longest wavelength, travels fastest in glass and is deviated the least. Consequently, light of different colours present in white light follow different paths inside the glass prism and then strike the second surface of the prism. On the second surface, only refraction takes place (from glass to air), and different colours are further deviated through different angles. Here again, violet is deviated the most and red the least. As a result, the colours get further separated on refraction at the second surface. The light emerging out of the prism, thus, has different colours that spread out to form a spectrum on the screen.<\/p>\n\n\n\n<p><strong>2. Explain briefly, with the help of a neat labelled diagram, how does white light get dispersed by a prism. On which surface of a prism, there is both dispersion and deviation of light, and on which surface of the prism, there is only deviation of light?<\/strong><br><strong><br>Answer<\/strong>: When white light from a source passes through a slit and then falls on a prism, it gets dispersed into its constituent colours. Newton allowed white light from the Sun to enter a dark room through a small aperture and placed a glass prism in its path. The light emerging out of the prism was received on a white screen, and a coloured patch like a rainbow, termed a spectrum, was obtained. Starting from the side of the base of the prism, the order of colours is Violet, Indigo, Blue, Green, Yellow, Orange, and Red (VIBGYOR). This shows that white light is a mixture of these colours.<\/p>\n\n\n\n<p>Dispersion of white light occurs only at the first surface of a prism. Deviation of light rays occurs at both the surfaces of a prism. Therefore, at the first surface of the prism, there is both dispersion and deviation of light. At the second surface of the prism, there is only deviation of light (as the colours are already separated and undergo refraction).<\/p>\n\n\n\n<img decoding=\"async\" src=\"https:\/\/blogger.googleusercontent.com\/img\/b\/R29vZ2xl\/AVvXsEiGnOxf8Dk31vdm0E3le-GVCwKLc-AVloKcKcuawG6ZlLmhAwMgQl7p5TI6z4okyE4hb7-4cZ5W4fV4kBBkV6z-JoaIa4CVUYxphdnUvBClAxQs53W2NQ8RTJNP7bRgLIayRK_CgDsbjD3WPXT2dBj0XKtz7JxvKdFuJ4hl3kogXZKsQNg04cOZGupRksR1\/s1200\/6.2.png\" style=\"max-width: 100%;height: auto\" \/>\n\n\n\n<p><strong>3. The diagram shown below shows the path taken by a narrow beam of yellow monochromatic light passing through an equiangular glass prism. If the yellow light is replaced by a narrow beam of white light incident at the same angle, draw another diagram to show the passage of white light through the prism and label it to show the effect of the prism on the white light.<\/strong><\/p>\n\n\n\n<img decoding=\"async\" src=\"https:\/\/blogger.googleusercontent.com\/img\/b\/R29vZ2xl\/AVvXsEgnlmoNzCaQY56PBTfux0kl84jbhr9ryU0zfo7NpvgdbtTgcl_iTCjfGpTkZA4OHxCEv5fnKYzMvxTViNj0tjDauOmQu1elBvEZ9cxwinkg0U7XfVi-BPyYJkoWb3f3yYtwrraWwEXXWc76IyKaDg5dL0n7KjXexq_0BxzQ126xy9DqOz3ShwGtRsd5FOA\/s900\/6.8.png\" style=\"max-width: 100%;height: auto\" \/>\n\n\n\n<p><strong>Answer<\/strong>: If the yellow light is replaced by a narrow beam of white light incident at the same angle, the white light will undergo dispersion as it passes through the equiangular glass prism. The diagram would show the incident white light ray striking the first face of the prism. Inside the prism, this ray would split into the band of seven colours (VIBGYOR), with violet light deviating the most towards the base and red light deviating the least. These coloured rays would then emerge from the second face of the prism, further deviated and separated, forming a spectrum on a screen if placed.<\/p>\n\n\n\n<img decoding=\"async\" src=\"https:\/\/blogger.googleusercontent.com\/img\/b\/R29vZ2xl\/AVvXsEiGnOxf8Dk31vdm0E3le-GVCwKLc-AVloKcKcuawG6ZlLmhAwMgQl7p5TI6z4okyE4hb7-4cZ5W4fV4kBBkV6z-JoaIa4CVUYxphdnUvBClAxQs53W2NQ8RTJNP7bRgLIayRK_CgDsbjD3WPXT2dBj0XKtz7JxvKdFuJ4hl3kogXZKsQNg04cOZGupRksR1\/s1200\/6.2.png\" style=\"max-width: 100%;height: auto\" \/>\n\n\n\n<p><strong>4. Figure shows a thin beam of white light from a source S striking on one face of a prism.<\/strong><\/p>\n\n\n\n<img decoding=\"async\" src=\"https:\/\/blogger.googleusercontent.com\/img\/b\/R29vZ2xl\/AVvXsEhWeTJmU7MZK0xCexNx05aUeXRiiSDRjl5pp6cXCFiEAn8MeH7zHII68pCg-o3fQJ1T-mTw79yU-6rMPLA6-bXQFTnElzfNRoXhAtkEn3fDoVW65-pG3k-pBECuReh0n5TNy08zQgbtcnqVebpbXTAS6xCKMMBJtx6MOac89YhpE2R2FiTPGj13VSl8V7M\/s900\/6.9.png\" style=\"max-width: 100%;height: auto\" \/>\n\n\n\n<p><strong>(a) Complete the diagram to show the effect of the prism on the beam and to show what is seen on the screen.<br>(b) If a slit is placed in between the prism and the screen to pass only the light of green colour, what will you then observe on the screen?<br>(c) What conclusion do you draw from the observation in part (b) above?<\/strong><br><strong><br>Answer<\/strong>: (a) The completed diagram would show the incident white light from source S striking one face of the prism. The white light would refract and disperse at this first surface, splitting into its constituent colours (VIBGYOR) inside the prism, with violet bent most and red least. These coloured rays would then refract again at the second surface as they emerge from the prism, diverging further. On the screen, a spectrum of colours (Violet, Indigo, Blue, Green, Yellow, Orange, Red) would be seen, with violet at the bottom (towards the base of the prism if oriented typically) and red at the top.<\/p>\n\n\n\n<img decoding=\"async\" src=\"https:\/\/blogger.googleusercontent.com\/img\/b\/R29vZ2xl\/AVvXsEiGnOxf8Dk31vdm0E3le-GVCwKLc-AVloKcKcuawG6ZlLmhAwMgQl7p5TI6z4okyE4hb7-4cZ5W4fV4kBBkV6z-JoaIa4CVUYxphdnUvBClAxQs53W2NQ8RTJNP7bRgLIayRK_CgDsbjD3WPXT2dBj0XKtz7JxvKdFuJ4hl3kogXZKsQNg04cOZGupRksR1\/s1200\/6.2.png\" style=\"max-width: 100%;height: auto\" \/>\n\n\n\n<p>(b) If a slit is placed in between the prism and the screen to pass only the light of green colour, then only a patch of green light will be observed on the screen.<\/p>\n\n\n\n<p>(c) The conclusion drawn from the observation in part (b) is that white light is composed of various colours (it is polychromatic), and green light is one of its constituent colours which was separated by the prism and isolated by the slit.<\/p>\n\n\n\n<p><strong>5. (a) A beam of monochromatic light undergoes minimum deviation through an equiangular prism. How does the beam pass through the prism with respect to its base?<br>(b) If white light is used in the same way as in part (a) above, what change do you expect in the emergent beam?<br>(c) What conclusion do you draw about the nature of white light in part (b)?<\/strong><br><strong><br>Answer<\/strong>: (a) When a beam of monochromatic light undergoes minimum deviation through an equiangular prism, the beam passes parallel to the base of the prism.<\/p>\n\n\n\n<p>(b) If white light is used in the same way as in part (a) above, the emergent beam will split into its constituent colours, i.e., a spectrum is formed.<\/p>\n\n\n\n<p>(c) The conclusion drawn about the nature of white light in part (b) is that white light is polychromatic.<\/p>\n\n\n\n<h5 class=\"wp-block-heading\" id=\"Numerical_Questions\"><strong>Numerical Questions<\/strong><\/h5>\n\n\n\n<p><strong>1. Calculate the frequency of yellow light of wavelength 550 nm. The speed of light is 3 \u00d7 10\u2078 m s\u207b\u00b9.<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong><\/p>\n\n\n\n<p>First, let&#8217;s list the given data in a table.<\/p>\n\n\n\n<p><strong>Given Data:<\/strong><br><\/p>\n\n\n\n<figure class=\"wp-block-table is-style-stripes\"><table class=\"has-fixed-layout\"><thead><tr><th>Quantity<\/th><th>Symbol<\/th><th>Value<\/th><\/tr><\/thead><tbody><tr><td>\u03bb<\/td><td>\u03bb<\/td><td>550 nm<\/td><\/tr><tr><td>c<\/td><td>c<\/td><td>3\u00d710\u2078 m\/s<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p>The relationship between the speed of light (c), frequency (\u03bd), and wavelength (\u03bb) is given by the formula:<br>c = \u03bd \u00d7 \u03bb<\/p>\n\n\n\n<p>To find the frequency (\u03bd), we can rearrange the formula as:<br>\u03bd = c \/ \u03bb<\/p>\n\n\n\n<p><strong>Step 1: Convert the wavelength to meters (m).<\/strong><br>We need to ensure all units are consistent. Since the speed of light is in meters per second (m s\u207b\u00b9), we must convert the wavelength from nanometers (nm) to meters (m).<\/p>\n\n\n\n<p>We know that 1 nm = 10\u207b\u2079 m.<br>Therefore, \u03bb = 550 nm = 550 \u00d7 10\u207b\u2079 m.<\/p>\n\n\n\n<p><strong>Step 2: Substitute the values into the formula and calculate the frequency.<\/strong><\/p>\n\n\n\n<p>\u03bd = c \/ \u03bb<br>=&gt; \u03bd = (3 \u00d7 10\u2078 m s\u207b\u00b9) \/ (550 \u00d7 10\u207b\u2079 m)<br>=&gt; \u03bd = (3 \/ 550) \u00d7 (10\u2078 \/ 10\u207b\u2079) s\u207b\u00b9<br>=&gt; \u03bd = 0.005454&#8230; \u00d7 10\u2078\u207b\u207d\u207b\u2079\u207e s\u207b\u00b9<br>=&gt; \u03bd = 0.005454&#8230; \u00d7 10\u00b9\u2077 s\u207b\u00b9<\/p>\n\n\n\n<p><strong>Step 3: Express the result in standard scientific notation.<\/strong><br>To write the number in proper scientific notation, we move the decimal point three places to the right.<\/p>\n\n\n\n<p>=&gt; \u03bd = 5.454&#8230; \u00d7 10\u207b\u00b3 \u00d7 10\u00b9\u2077 s\u207b\u00b9<br>=&gt; \u03bd = 5.45 \u00d7 10\u00b9\u2074 s\u207b\u00b9<\/p>\n\n\n\n<p>The unit s\u207b\u00b9 is also known as Hertz (Hz).<\/p>\n\n\n\n<p>Thus, the frequency of the yellow light is 5.45 \u00d7 10\u00b9\u2074 Hz.<\/p>\n\n\n\n<p><strong>2. The frequency range of visible light is from 3.75 x 10\u00b9\u2074 Hz to 7.5 x 10\u00b9\u2074 Hz. Calculate its wavelength range. Take speed of light = 3 \u00d7 10\u2078 m s\u207b\u00b9.<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong> The relationship between the speed of light (c), frequency (\u03bd), and wavelength (\u03bb) is given by the formula:<br>\u03bb = c \/ \u03bd<\/p>\n\n\n\n<p>To find the wavelength range, we will calculate the corresponding wavelengths for the minimum and maximum given frequencies.<\/p>\n\n\n\n<p><strong>(i) Wavelength for the lower frequency limit (\u03bd\u2081 = 3.75 x 10\u00b9\u2074 Hz):<\/strong><\/p>\n\n\n\n<p>This frequency corresponds to the maximum wavelength (\u03bb_max) because wavelength is inversely proportional to frequency.<\/p>\n\n\n\n<p>Given:<br>Frequency (\u03bd\u2081) = 3.75 x 10\u00b9\u2074 Hz<br>Speed of light (c) = 3 x 10\u2078 m s\u207b\u00b9<\/p>\n\n\n\n<p>Using the formula:<br>\u03bb_max = c \/ \u03bd\u2081<br>\u03bb_max = (3 x 10\u2078 m s\u207b\u00b9) \/ (3.75 x 10\u00b9\u2074 Hz)<br>=&gt; \u03bb_max = (3 \/ 3.75) x 10^(8 &#8211; 14) m<br>=&gt; \u03bb_max = 0.8 x 10\u207b\u2076 m<br>=&gt; \u03bb_max = 8 x 10\u207b\u2077 m<\/p>\n\n\n\n<p><strong>(ii) Wavelength for the higher frequency limit (\u03bd\u2082 = 7.5 x 10\u00b9\u2074 Hz):<\/strong><\/p>\n\n\n\n<p>This frequency corresponds to the minimum wavelength (\u03bb_min).<\/p>\n\n\n\n<p>Given:<br>Frequency (\u03bd\u2082) = 7.5 x 10\u00b9\u2074 Hz<br>Speed of light (c) = 3 x 10\u2078 m s\u207b\u00b9<\/p>\n\n\n\n<p>Using the formula:<br>\u03bb_min = c \/ \u03bd\u2082<br>\u03bb_min = (3 x 10\u2078 m s\u207b\u00b9) \/ (7.5 x 10\u00b9\u2074 Hz)<br>=&gt; \u03bb_min = (3 \/ 7.5) x 10^(8 &#8211; 14) m<br>=&gt; \u03bb_min = 0.4 x 10\u207b\u2076 m<br>=&gt; \u03bb_min = 4 x 10\u207b\u2077 m<\/p>\n\n\n\n<p>Thus, the wavelength range of visible light is from 4 x 10\u207b\u2077 m to 8 x 10\u207b\u2077 m.<\/p>\n\n\n\n<h4 class=\"wp-block-heading\" id=\"Very_Short_Answer_Type_Questions\">Exercise (B)<\/h4>\n\n\n\n<h5 class=\"wp-block-heading\" id=\"Very_Short_Answer_Type_Questions\"><strong>MCQ<\/strong><\/h5>\n\n\n\n<p><strong>1. When an electromagnetic wave passes from one medium to the other, which property remains unchanged?<\/strong><\/p>\n\n\n\n<p>(a) speed<br>(b) wavelength<br>(c) direction of travel<br>(d) frequency<\/p>\n\n\n\n<p><strong>Answer<\/strong>: (d) frequency<\/p>\n\n\n\n<p><strong>2. Two waves A and B have wavelengths 0-01 A\u00ba and 9000 A\u00b0 respectively. The waves A and B are :<\/strong><\/p>\n\n\n\n<p>(a) Gamma wave and ultraviolet wave respectively.<br>(b) Microwave and infrared wave respectively.<br>(c) Radiowave and gamma wave respectively.<br>(d) Gamma wave and infrared wave respectively.<\/p>\n\n\n\n<p><strong>Answer<\/strong>: (d) Gamma wave and infrared wave respectively.<\/p>\n\n\n\n<p><strong>3. The correct arrangement of the following radiations in an increasing order of their wavelengths is :<\/strong><strong><br><\/strong><strong>X-rays, infrared rays, gamma rays, microwaves.<\/strong><\/p>\n\n\n\n<p>(a) gamma rays, infrared rays, X-rays, microwaves<br>(b) gamma rays, X-rays, microwaves, infrared rays<br>(c) gamma rays, X-rays, infrared rays, microwaves<br>(d) X-rays, gamma rays, infrared rays, microwaves<\/p>\n\n\n\n<p><strong>Answer<\/strong>: (c) gamma rays, X-rays, infrared rays, microwaves<\/p>\n\n\n\n<p><strong>4. A radiation P is focused by a proper device on the bulb of a thermometer. Mercury in the thermometer shows a rapid increase. The radiation P is :<\/strong><\/p>\n\n\n\n<p>(a) infrared radiation<br>(b) visible light<br>(c) ultraviolet radiation<br>(d) X-rays.<\/p>\n\n\n\n<p><strong>Answer<\/strong>: (a) infrared radiation<\/p>\n\n\n\n<p><strong>5. The most energetic electromagnetic radiations are:<\/strong><\/p>\n\n\n\n<p>(a) microwaves<br>(b) ultraviolet waves<br>(c) X-rays<br>(d) gamma rays.<\/p>\n\n\n\n<p><strong>Answer<\/strong>: (d) gamma rays.<\/p>\n\n\n\n<p><strong>6. Column X shows the kinds of electromagnetic waves and column Y shows their applications.<\/strong><\/p>\n\n\n\n<p><strong>(A) Infrared rays (i) in remote-controlled gadgets<br>(B) Radio waves (ii) for transmission<br>(C) X-rays (iii) for detection of bone fractures<br>(D) Ultraviolet rays (iv) absorption by atmospheric ozone layer<br>Choose the correct pairing:<\/strong><\/p>\n\n\n\n<p>(a) A-(i) B-(ii) C-(iii) D-(iv)<br>(b) A-(iv) B-(iii) C-(ii) D-(i)<br>(c) A-(i) B-(ii) C-(iv) D-(iii)<br>(d) A-(iii) B-(ii) C-(i) D-(iv)<\/p>\n\n\n\n<p><strong>Answer<\/strong>: (a) A-(i) B-(ii) C-(iii) D-(iv)<\/p>\n\n\n\n<h5 class=\"wp-block-heading\" id=\"Very_Short_Answer_Type_Questions\"><strong>Very Short Answer Type Questions<\/strong><\/h5>\n\n\n\n<p><strong>1. (a) Arrange the following radiations in the order of their increasing wavelength : X-rays, infrared rays, radio waves, gamma rays, and micro waves.<\/strong><br><strong><br>Answer<\/strong>: The radiations in the order of their increasing wavelength are: gamma rays, X-rays, infrared rays, micro waves, and radio waves.<\/p>\n\n\n\n<p><strong>(b) Name the radiation which is used for satellite communication.<\/strong><br><strong><br>Answer<\/strong>: Micro waves are used for satellite communication.<\/p>\n\n\n\n<p><strong>2. A wave has a wavelength 10\u207b\u00b3 nm. (a) Name the wave. (b) State its one property different from light.<\/strong><br><strong><br>Answer<\/strong>: (a) The wave is a gamma ray.<br>(b) One property different from light is its strong penetrating power; for example, gamma rays can easily penetrate through thick metallic sheets.<\/p>\n\n\n\n<p><strong>3. (a) Name the high energetic invisible electromagnetic wave which helps in the study of the structure of crystals.<\/strong><br><strong><br>Answer<\/strong>: (a) The high energetic invisible electromagnetic wave which helps in the study of the structure of crystals is X-rays.<\/p>\n\n\n\n<p><strong>(b) State one more use of the wave named in part (a).<\/strong><br><strong><br>Answer<\/strong>: One more use of X-rays is for the detection of fracture in bones.<\/p>\n\n\n\n<p><strong>4. State the name and the range of wavelength of the invisible electromagnetic waves beyond the red end of the visible spectrum.<\/strong><br><strong><br>Answer<\/strong>: The invisible electromagnetic wave beyond the red end of the visible spectrum is Infrared radiation. Its wavelength range is from 8000 \u00c5 to 10\u2077 \u00c5 (or 800 nm to 1 mm).<\/p>\n\n\n\n<p><strong>5. Give the range of wavelength of the electromagnetic waves visible to us.<\/strong><br><strong><br>Answer<\/strong>: The range of wavelength of the electromagnetic waves visible to us is from 4000 \u00c5 to 8000 \u00c5 (or 400 nm to 800 nm).<\/p>\n\n\n\n<p><strong>6. Name the region just beyond (i) the red end, and (ii) the violet end, of the spectrum.<\/strong><br><strong><br>Answer<\/strong>: (i) The region just beyond the red end of the spectrum is the infrared spectrum.<br>(ii) The region just before the violet end of the visible spectrum is the ultraviolet spectrum.<\/p>\n\n\n\n<p><strong>7. Name the radiation which can be detected by (a) a thermopile (b) a solution of silver chloride.<\/strong><br><strong><br>Answer<\/strong>: (a) Infrared radiation can be detected by a thermopile.<br>(b) Ultraviolet radiation can be detected by a solution of silver chloride.<\/p>\n\n\n\n<p><strong>8. Name the radiations of wavelength just (a) longer than 8 x 10\u207b\u2077 m, (b) shorter than 4 x 10\u207b\u2077 m.<\/strong><br><strong><br>Answer<\/strong>: (a) The radiation of wavelength just longer than 8 x 10\u207b\u2077 m (800 nm) is infrared radiation.<br>(b) The radiation of wavelength just shorter than 4 x 10\u207b\u2077 m (400 nm) is ultraviolet radiation.<\/p>\n\n\n\n<p><strong>9. Name the material of prism required for obtaining the spectrum of (a) ultraviolet light, (b) infrared radiations.<\/strong><br><strong><br>Answer<\/strong>: (a) The material of prism required for obtaining the spectrum of ultraviolet light is quartz.<br>(b) The material of prism required for obtaining the spectrum of infrared radiations is rock-salt.<\/p>\n\n\n\n<p><strong>10. Name the radiations which are absorbed by the green house gases in the earth&#8217;s atmosphere.<\/strong><br><strong><br>Answer<\/strong>: The radiations which are absorbed by the green house gases such as carbon-dioxide in the earth&#8217;s atmosphere are the low energy infrared radiations.<\/p>\n\n\n\n<h5 class=\"wp-block-heading\" id=\"Very_Short_Answer_Type_Questions\"><strong>Short Answer Type Questions<\/strong><\/h5>\n\n\n\n<p><strong>1. (a) Give a list of at least five radiations, in the order of their increasing wavelength, which make up the complete electromagnetic spectrum.<\/strong><br><strong><br>Answer<\/strong>: (a) A list of five radiations in the order of their increasing wavelength which make up the complete electromagnetic spectrum is:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Gamma rays<\/li>\n\n\n\n<li>X-rays<\/li>\n\n\n\n<li>Ultraviolet rays<\/li>\n\n\n\n<li>Visible light<\/li>\n\n\n\n<li>Infrared radiations<\/li>\n<\/ul>\n\n\n\n<p><strong>(b) Name the radiation mentioned by you in part (a) which has the highest penetrating power.<\/strong><br><strong><br>Answer<\/strong>:(b) The radiation mentioned in part (a) which has the highest penetrating power is Gamma rays.<\/p>\n\n\n\n<p><strong>2. A wave has wavelength 50 \u00c5. (a) Name the wave. (b) State its speed in vacuum. (c) State its one use.<\/strong><br><strong><br>Answer<\/strong>: (a) The wave with wavelength 50 \u00c5 is an X-ray.<br>(b) Its speed in vacuum is 3 \u00d7 10\u2078 m s\u207b\u00b9.<br>(c) One use of X-rays is in the detection of fracture in bones.<\/p>\n\n\n\n<p><strong>3. Name three radiations and their wavelength range which are invisible and beyond the violet end of the visible spectrum.<\/strong><br><strong><br>Answer<\/strong>: Three radiations which are invisible and beyond the violet end (i.e., have shorter wavelengths) of the visible spectrum are:<\/p>\n\n\n\n<p>(i) Ultraviolet rays: wavelength range 100 \u00c5 to 4000 \u00c5 (or 10 nm to 400 nm).<br>(ii) X-rays: wavelength range 0.1 \u00c5 to 100 \u00c5 (or 0.01 nm to 10 nm).<br>(iii) Gamma rays: wavelength shorter than 0.1 \u00c5 (or \u03bb &lt; 0.01 nm).<\/p>\n\n\n\n<p><strong>4. What do you understand by the invisible spectrum ?<\/strong><br><strong><br>Answer<\/strong>: The spectrum of radiation from the sun extends on either side beyond the violet and red extremes to which our eyes do not respond. This part of the spectrum beyond the red extreme and the violet extreme is called the invisible spectrum. The portion of electromagnetic waves other than the visible part, do not excite our retina to produce the sensation of vision.<\/p>\n\n\n\n<p><strong>5. State the approximate range of wavelength associated with (a) ultraviolet rays, (b) visible light, and (c) infrared rays.<\/strong><br><strong><br>Answer<\/strong>: The approximate ranges of wavelength are:<\/p>\n\n\n\n<p>(a) Ultraviolet rays: 100 \u00c5 to 4000 \u00c5 (or 10 nm to 400 nm).<br>(b) Visible light: 4000 \u00c5 to 8000 \u00c5 (or 400 nm to 800 nm).<br>(c) Infrared rays: 8000 \u00c5 to 10\u2077 \u00c5 (or 800 nm to 1 mm).<\/p>\n\n\n\n<p><strong>6. Name two electromagnetic waves of wavelength smaller than that of violet light. State one use of each.<\/strong><br><strong><br>Answer<\/strong>: Two electromagnetic waves of wavelength smaller than that of violet light are:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Ultraviolet rays: One use is for sterilising air and surgical equipments.<\/li>\n\n\n\n<li>X-rays: One use is for the detection of fracture in bones and teeth.<\/li>\n<\/ul>\n\n\n\n<p><strong>7. Give one use each of (a) microwaves, (b) ultraviolet radiations, (c) infrared radiations, and (d) gamma rays.<\/strong><br><strong><br>Answer<\/strong>: (a) Microwaves: Used for satellite communication.<br>(b) Ultraviolet radiations: Used for detecting the purity of gems, eggs, ghee, etc.<br>(c) Infrared radiations: Used in remote control of television and other gadgets.<br>(d) Gamma rays: Used in medical science to kill cancer cells (i.e., radio therapy).<\/p>\n\n\n\n<p><strong>8. Name the waves (a) of lowest wavelength, (b) used for taking photographs in dark, (c) produced by changes in the nucleus of an atom, (d) of wavelength nearly 0.1 nm.<\/strong><br><strong><br>Answer<\/strong>: (a) The waves of lowest wavelength are gamma rays.<br>(b) The waves used for taking photographs in dark are infrared radiations.<br>(c) The waves produced by changes in the nucleus of an atom are gamma rays.<br>(d) The waves of wavelength nearly 0.1 nm (1 \u00c5) are X-rays.<\/p>\n\n\n\n<p><strong>9. Two waves A and B have wavelength 0.01 \u00c5 and 9000 \u00c5 respectively. (a) Name the two waves. (b) Compare the speeds of these waves when they travel in vacuum.<\/strong><br><strong><br>Answer<\/strong>: (a) Wave A with wavelength 0.01 \u00c5 is a gamma ray. Wave B with wavelength 9000 \u00c5 is an infrared radiation.<br>(b) When they travel in vacuum, both waves A and B have the same speed, which is 3 \u00d7 10\u2078 m s\u207b\u00b9. Their speeds are equal, so the ratio of their speeds is 1:1.<\/p>\n\n\n\n<p><strong>10. Name two sources, each of infrared radiations and ultraviolet radiations.<\/strong><br><strong><br>Answer<\/strong>: Two sources of infrared radiations are:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>A heated iron ball.<\/li>\n\n\n\n<li>The Sun.<\/li>\n<\/ul>\n\n\n\n<p>Two sources of ultraviolet radiations are:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>An electric arc.<\/li>\n\n\n\n<li>A mercury vapour lamp.<\/li>\n<\/ul>\n\n\n\n<p><strong>11. Name two properties of ultraviolet radiations which are similar to visible light and two which differ from visible light.<\/strong><br><strong><br>Answer<\/strong>:Two properties of ultraviolet radiations similar to visible light are:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>They travel in a straight line with a speed of 3 \u00d7 10\u2078 m s\u207b\u00b9 in air (or vacuum).<\/li>\n\n\n\n<li>They obey the laws of reflection and refraction.<\/li>\n<\/ul>\n\n\n\n<p>Two properties of ultraviolet radiations different from visible light are:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Ultraviolet radiations are invisible, while visible light is visible.<\/li>\n\n\n\n<li>Ultraviolet radiations can pass through quartz but are absorbed by glass, whereas visible light passes through glass.<\/li>\n<\/ul>\n\n\n\n<p><strong>12. Mention two properties of infrared radiations which are similar to visible light and two which differ from visible light.<\/strong><br><strong><br>Answer<\/strong>:Two properties of infrared radiations similar to visible light are:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>They travel in straight lines with the same speed as light (3 \u00d7 10\u2078 m s\u207b\u00b9) in vacuum (or air).<\/li>\n\n\n\n<li>They obey the laws of reflection and refraction.<\/li>\n<\/ul>\n\n\n\n<p>Two properties of infrared radiations different from visible light are:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Infrared radiations are invisible, while visible light is visible.<\/li>\n\n\n\n<li>Infrared radiations produce a strong heating effect, while visible light produces a slight heating effect. (Alternatively, infrared radiations are absorbed by glass, but they pass through rock-salt, while visible light passes through glass).<\/li>\n<\/ul>\n\n\n\n<p><strong>13. State one harmful effect each of the (a) ultraviolet and (b) infrared radiations.<\/strong><br><strong><br>Answer<\/strong>:(a) One harmful effect of ultraviolet radiations is that they can cause health hazards like skin cancer if the human body is exposed to them for a long period.<br>(b) One harmful effect of infrared radiations is that a high dose may cause skin burns.<\/p>\n\n\n\n<p><strong>14. Give reason for the following :<\/strong><\/p>\n\n\n\n<p><strong>(i) Infrared radiations are used for photography in fog.<\/strong><br><strong><br>Answer<\/strong>: Infrared radiations are used for photography in fog because they are scattered less by the earth&#8217;s atmosphere due to their long wavelength. Hence, they are able to penetrate deep inside the atmosphere even in fog and are not much scattered, so they can penetrate appreciably through it.<\/p>\n\n\n\n<p><strong>(ii) Infrared radiations are used for signals during war.<\/strong><br><strong><br>Answer<\/strong>: Infrared radiations are used for signals during war as they are not visible and they are not absorbed much in the medium.<\/p>\n\n\n\n<p><strong>(iii) The photographic darkrooms are provided with infrared lamps.<\/strong><br><strong><br>Answer<\/strong>: The photographic darkrooms are provided with infrared lamps because these lamps provide some visibility without affecting the ordinary photographic film, as infrared radiations do not affect it.<\/p>\n\n\n\n<p><strong>(iv) A rock salt prism is used instead of a glass prism to obtain the infrared spectrum.<\/strong><br><strong><br>Answer<\/strong>: A rock-salt prism is used instead of a glass prism to obtain the infrared spectrum because a rock-salt prism does not absorb infrared radiations, whereas a glass prism absorbs them.<\/p>\n\n\n\n<p><strong>(v) A quartz prism is required for obtaining the spectrum of ultraviolet light.<\/strong><br><strong><br>Answer<\/strong>: A quartz prism is required for obtaining the spectrum of ultraviolet light because ultraviolet radiations can pass through quartz, but they are absorbed by glass.<\/p>\n\n\n\n<p><strong>(vi) Ultraviolet bulbs have a quartz envelope instead of glass.<\/strong><br><strong><br>Answer<\/strong>: Ultraviolet bulbs have an envelope made of quartz instead of glass because quartz transmits ultraviolet radiations while glass absorbs them.<\/p>\n\n\n\n<p><strong>15. Radiations from the sun fall on a prism and suffer dispersion. Three thermometers are kept on which the radiations after dispersion are made to fall as shown in the figure given alongside. Which thermometer would show a higher reading? Give a reason for your answer.<\/strong><br><strong><br>Answer<\/strong>: Thermometer A (placed in the red region) would show a higher reading compared to thermometers B (in indigo) and C (in violet).<\/p>\n\n\n\n<p>The reason is that although there is a slow rise in temperature from the violet end towards the red end of the visible spectrum, the heating effect is more pronounced towards the red end. Furthermore, the region just beyond the red extreme (infrared radiations) has the strongest heating effect. Since thermometer A is in the red region, it is closest to this high-heat infrared region and receives more heat energy from the visible spectrum compared to indigo and violet.<\/p>\n\n\n\n<img decoding=\"async\" src=\"https:\/\/blogger.googleusercontent.com\/img\/b\/R29vZ2xl\/AVvXsEhMdL7o53aPSDpuvKDzViMTyHlT-kA01THE_GaiIYV_ElZSbc4IRJYNcyaoLzrSyGfXbjm9fNfGrfY_LWVrVqRQZIn4Av8ebN8s7e465BDXgSlcgcde5vM5rF-NgrjElmdQI9wgxPtKNRhK9KtV-LvzF7x_E5lHiWvS4_xWYj4nAbnpe4Nx0rXFv0s3f77i\/s1200\/6.3.png\" style=\"max-width: 100%;height: auto\" \/>\n\n\n\n<h5 class=\"wp-block-heading\" id=\"Very_Short_Answer_Type_Questions\"><strong>Long Answer Type Questions<\/strong><\/h5>\n\n\n\n<p><strong>1. What are infrared radiations ? How are they detected ? State one use of these radiations.<\/strong><br><strong><br>Answer<\/strong>: Infrared radiations are electromagnetic waves of wavelength in the range of 8000 \u00c5 to 10\u2077 \u00c5 (or 800 nm to 1 mm). These are radiations which produce a strong heating effect and are part of the spectrum beyond the red extreme of visible light.<\/p>\n\n\n\n<p>Infrared radiations are detected by their heating property.<\/p>\n\n\n\n<p>(i) If a thermometer having its bulb blackened is moved from the violet end towards the red end of the spectrum and then beyond the red extreme, a rapid rise in temperature is noticed in the region beyond red, indicating the presence of infrared radiations.<br>(ii) A thermopile can also be used to detect heat radiations; a galvanometer connected to the thermopile shows deflection when infrared radiations fall on it.<\/p>\n\n\n\n<p>One use of infrared radiations is that they are used for therapeutic purposes by doctors. Another use is in photography at night and also in mist and fog because they are not much scattered.<\/p>\n\n\n\n<p><strong>2. What are ultraviolet radiations ? How are they detected ? State one use of these radiations.<\/strong><br><strong><br>Answer<\/strong>: Ultraviolet radiations are electromagnetic radiations of wavelength from 100 \u00c5 to 4000 \u00c5 (or 10 nm to 400 nm). These radiations are found just beyond the violet extreme of the visible part of the spectrum and are chemically more active than visible light.<\/p>\n\n\n\n<p>Ultraviolet radiations can be detected in a few ways:<\/p>\n\n\n\n<p>(i) When a silver-chloride solution is exposed to electromagnetic waves, starting from the red to the violet end and then beyond it, the solution remains almost unaffected in the visible region. However, just beyond the violet end, the solution first turns violet and then finally it becomes dark brown (or black), indicating the presence of ultraviolet radiations.<br>(ii) They can be detected by their chemical activity on dyes and photographic plates, as they strongly affect a photographic plate.<br>(iii) They produce fluorescence on striking a zinc-sulphide screen.<\/p>\n\n\n\n<p>One use of ultraviolet radiations is for sterilising air and surgical equipments. Another use is for detecting the purity of gems, eggs, ghee, etc.<\/p>\n\n\n\n<h5 class=\"wp-block-heading\" id=\"Very_Short_Answer_Type_Questions\"><strong>Numericals<\/strong><\/h5>\n\n\n\n<p><strong>1. An electromagnetic wave has a frequency of 500 MHz and a wavelength of 60 cm. (a) Calculate the speed of the wave. (b) Name the medium through which it is travelling.<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong> <strong>(a) Calculate the speed of the wave.<\/strong><\/p>\n\n\n\n<p>First, we list the given values and convert them to standard SI units.<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Frequency (f) = 500 MHz<br>=&gt; f = 500 \u00d7 10\u2076 Hz (Since 1 MHz = 10\u2076 Hz)<\/li>\n\n\n\n<li>Wavelength (\u03bb) = 60 cm<br>=&gt; \u03bb = 60 \/ 100 m = 0.6 m (Since 1 m = 100 cm)<\/li>\n<\/ul>\n\n\n\n<p>Now, we use the formula for the speed of a wave:<br>Speed (v) = Frequency (f) \u00d7 Wavelength (\u03bb)<\/p>\n\n\n\n<p>v = (500 \u00d7 10\u2076) \u00d7 0.6<br>=&gt; v = 300 \u00d7 10\u2076<br>=&gt; v = 3 \u00d7 10\u00b2 \u00d7 10\u2076<br>=&gt; v = 3 \u00d7 10\u2078 m\/s.<\/p>\n\n\n\n<p>The speed of the wave is 3 \u00d7 10\u2078 m\/s.<\/p>\n\n\n\n<p><strong>(b) Name the medium through which it is travelling.<\/strong><\/p>\n\n\n\n<p>The calculated speed of the wave is 3 \u00d7 10\u2078 m\/s. This is the well-known value for the speed of light (and all electromagnetic waves) in a vacuum.<\/p>\n\n\n\n<p>Therefore, the medium through which the wave is travelling is a <strong>vacuum<\/strong> (or air, as the speed in air is approximately the same).<\/p>\n\n\n\n<p><strong>2. The wavelength of X-rays is 0.01 \u00c5. Calculate its frequency. State the assumption made, if any.<\/strong><\/p>\n\n\n\n<p><strong>Answer:<\/strong><\/p>\n\n\n\n<p><strong>Given:<\/strong><br>Wavelength of X-rays (\u03bb) = 0.01 \u00c5<\/p>\n\n\n\n<p><strong>To find:<\/strong><br>Frequency (\u03bd)<\/p>\n\n\n\n<p><strong>Formula:<\/strong><br>The relationship between the speed of an electromagnetic wave (c), its frequency (\u03bd), and its wavelength (\u03bb) is given by the formula:<br>c = \u03bd\u03bb<br>Therefore, frequency (\u03bd) can be calculated as:<br>\u03bd = c \/ \u03bb<\/p>\n\n\n\n<p><strong>Step 1: Convert the wavelength from Angstroms (\u00c5) to meters (m).<\/strong><br>We know that 1 \u00c5 = 10\u207b\u00b9\u2070 m.<br>Wavelength (\u03bb) = 0.01 \u00c5<br>=&gt; \u03bb = 0.01 \u00d7 10\u207b\u00b9\u2070 m<br>=&gt; \u03bb = 1 \u00d7 10\u207b\u00b2 \u00d7 10\u207b\u00b9\u2070 m<br>=&gt; \u03bb = 1 \u00d7 10\u207b\u00b9\u00b2 m<\/p>\n\n\n\n<p><strong>Step 2: Calculate the frequency using the formula.<\/strong><br>We will use the value for the speed of light in a vacuum, c = 3 \u00d7 10\u2078 m\/s.<\/p>\n\n\n\n<p>\u03bd = c \/ \u03bb<br>=&gt; \u03bd = (3 \u00d7 10\u2078 m\/s) \/ (1 \u00d7 10\u207b\u00b9\u00b2 m)<br>=&gt; \u03bd = 3 \u00d7 10\u207d\u2078\u207b\u207d\u207b\u00b9\u00b2\u207e\u207e Hz<br>=&gt; \u03bd = 3 \u00d7 10\u207d\u2078\u207a\u00b9\u00b2\u207e Hz<br>=&gt; \u03bd = 3 \u00d7 10\u00b2\u2070 Hz<\/p>\n\n\n\n<p>Thus, the frequency of the X-rays is 3 \u00d7 10\u00b2\u2070 Hz.<\/p>\n\n\n\n<p><strong>Assumption made:<\/strong><br>The assumption is that the X-rays are traveling in a vacuum (or air), so their speed is taken as the speed of light in a vacuum, which is c = 3 \u00d7 10\u2078 m\/s.<\/p>\n\n\n\n<h4 class=\"wp-block-heading\" id=\"Very_Short_Answer_Type_Questions\"><strong>Exercise (C)<\/strong><\/h4>\n\n\n\n<h5 class=\"wp-block-heading\" id=\"Very_Short_Answer_Type_Questions\"><strong>MCQ<\/strong><\/h5>\n\n\n\n<p><strong>1. For scattering of light, the necessary condition is that the size of air molecules should be &#8230;&#8230;&#8230;&#8230; the wavelength of incident light.<\/strong><\/p>\n\n\n\n<p>(a) smaller than<br>(b) greater than<br>(c) equal to<br>(d) both (b) and (c)<\/p>\n\n\n\n<p><strong>Answer:<\/strong> (a) smaller than<\/p>\n\n\n\n<p><strong>2. The intensity of scattered light is related to its wavelength as:<\/strong><\/p>\n\n\n\n<p>(a) I \u221d \u03bb\u2074<br>(b) I \u221d \u03bb\u00b2<br>(c) I \u221d 1\/\u03bb\u2074<br>(d) I \u221d 1\/\u03bb\u00b2<\/p>\n\n\n\n<p><strong>Answer:<\/strong> (c) I \u221d 1\/\u03bb\u2074<\/p>\n\n\n\n<p><strong>3. In the white light of the sun, the maximum scattering by the air molecules present in the earth&#8217;s atmosphere is for :<\/strong><\/p>\n\n\n\n<p>(a) red colour<br>(b) yellow colour<br>(c) green colour<br>(d) blue colour<\/p>\n\n\n\n<p><strong>Answer:<\/strong> (d) blue colour<\/p>\n\n\n\n<p><strong>4. Violet light is scattered nearly &#8230;&#8230; times more than the red light.<\/strong><\/p>\n\n\n\n<p>(a) 4<br>(b) 8<br>(c) 16<br>(d) 9<\/p>\n\n\n\n<p><strong>Answer:<\/strong> (c) 16<\/p>\n\n\n\n<p><strong>5. To an astronaut in a spaceship, the earth appears:<\/strong><\/p>\n\n\n\n<p>(a) white<br>(b) red<br>(c) blue<br>(d) black<\/p>\n\n\n\n<p><strong>Answer:<\/strong> (c) blue<\/p>\n\n\n\n<p><strong>6. The red colour of the sun at sunrise and sunset is due to which of the following phenomenon ?<\/strong><\/p>\n\n\n\n<p>(a) deviation<br>(b) dispersion<br>(c) both deviation and dispersion<br>(d) scattering<\/p>\n\n\n\n<p><strong>Answer:<\/strong> (d) scattering<\/p>\n\n\n\n<p><strong>7. The sky in a direction other than that of the sun appears &#8230;&#8230;&#8230;&#8230; to an observer on the moon.<\/strong><\/p>\n\n\n\n<p>(a) red<br>(b) blue<br>(c) orange<br>(d) black<\/p>\n\n\n\n<p><strong>Answer:<\/strong> (d) black<\/p>\n\n\n\n<p><strong>8. The danger signal is red because its wavelength is the &#8230;&#8230;&#8230;&#8230; and therefore it gets &#8230;&#8230;&#8230;&#8230; the least.<\/strong><\/p>\n\n\n\n<p>(a) least, deviated<br>(b) longest, scattered<br>(c) longest, dispersed<br>(d) least, scattered<\/p>\n\n\n\n<p><strong>Answer:<\/strong> (b) longest, scattered<\/p>\n\n\n\n<h5 class=\"wp-block-heading\" id=\"Very_Short_Answer_Type_Questions\"><strong>Very Short Answer Type Questions<\/strong><\/h5>\n\n\n\n<p><strong>1. When sunlight enters the earth&#8217;s atmosphere, state which colour of light is scattered (i) the most, and (ii) the least.<\/strong><br><strong><br>Answer<\/strong>: (i) Violet light is scattered the most.<br>(ii) Red light is scattered the least.<\/p>\n\n\n\n<p><strong>2. A beam of blue, green and yellow light passes through the earth&#8217;s atmosphere. Name the colour which is scattered (a) the least, (b) the most.<\/strong><br><strong><br>Answer<\/strong>: (a) Yellow light is scattered the least among the given colours.<br>(b) Blue light is scattered the most among the given colours.<\/p>\n\n\n\n<h5 class=\"wp-block-heading\" id=\"Very_Short_Answer_Type_Questions\"><strong>Short Answer Type Questions<\/strong><\/h5>\n\n\n\n<p><strong>1. What is meant by scattering of light ?<\/strong><br><strong><br>Answer<\/strong>: Scattering is the process of absorption and then re-emission of light energy by the dust particles and air molecules present in the atmosphere.<\/p>\n\n\n\n<p><strong>2. How does the intensity of scattered light depend on the wavelength of incident light ? State the condition when this dependence hold.<\/strong><br><strong><br>Answer<\/strong>: The intensity of the scattered light is found to be inversely proportional to the fourth power of the wavelength of light (i.e., I \u221d 1\/\u03bb\u2074). This dependence holds when the scattering is by air molecules of size smaller than the wavelength of the incident light.<\/p>\n\n\n\n<p><strong>3. Which colour of white light is scattered the least ? Give reason.<\/strong><br><strong><br>Answer<\/strong>: Red colour of white light is scattered the least because the wavelength of red light is longest and the intensity of scattered light I \u221d 1\/\u03bb\u2074.<\/p>\n\n\n\n<p><strong>4. The danger signal is red. Why ?<\/strong><br><strong><br>Answer<\/strong>: The danger signal is red because in visible light, the wavelength of red light is longest, therefore the light of red colour is scattered the least by the air molecules of the atmosphere. Hence, the light of red colour as compared to the light of other colours can penetrate to a longer distance without becoming weak. Thus, red light can be seen from the farthest distance in comparison to the light of other colours having the same intensity, so that the signal may be visible from a far distance even in fog, etc.<\/p>\n\n\n\n<p><strong>5. How would the sky appear when seen from the space (or moon) ? Give reason for your answer.<\/strong><br><strong><br>Answer<\/strong>: When seen from space or the moon, the sky would appear black. This is because there is no atmosphere in space or on the Moon, so no scattered sunlight reaches the observer&#8217;s eyes, except for light reaching directly from the Sun. Thus, the sky in a direction other than that of the Sun will appear black.<\/p>\n\n\n\n<p><strong>6. What characteristic property of light is responsible for the blue colour of sky ?<\/strong><br><strong><br>Answer<\/strong>: The characteristic property of light responsible for the blue colour of the sky is scattering.<\/p>\n\n\n\n<p><strong>7. The colour of sky, in direction other than of the Sun, is blue. Explain.<\/strong><br><strong><br>Answer<\/strong>: As light travels through the atmosphere, it gets scattered in different directions by the air molecules present in its path. The blue (or violet) light due to its short wavelength is scattered more as compared to the red light of long wavelength. Thus, the light reaching our eye directly from the Sun is rich in red colour, while the light reaching our eye from all other directions is the scattered blue light. Therefore the sky in a direction, other than the direction of the sun, is seen blue.<\/p>\n\n\n\n<p><strong>8. Why does the Sun appear red at sunrise and sunset?<\/strong><br><strong><br>Answer<\/strong>: At the time of sunrise and sunset, light from the Sun has to travel the longest distance of atmosphere to reach the observer. Since the blue light of short wavelength is scattered more, much of it is lost, while the red light of long wavelength is scattered a little, so it is not much lost. Thus, blue light is almost absent in sunlight reaching the observer and only red (white &#8211; blue = red) light reaches us. As a result, the Sun and the region nearby it, is seen red.<\/p>\n\n\n\n<p><strong>9. The sky at noon appears white. Give reason.<\/strong><br><strong><br>Answer<\/strong>: At noon, the Sun is directly above our head, so we get light rays directly from the Sun after travelling the shortest distance, without much scattering of any particular colour. Hence, the sky is seen white.<\/p>\n\n\n\n<p><strong>10. The clouds are seen white. Explain.<\/strong><br><strong><br>Answer<\/strong>: The clouds are nearer to the earth&#8217;s surface and they contain dust particles and aggregates of water molecules of size bigger than the wavelength of visible light. Therefore, the dust particles and tiny ice particles present in clouds scatter all colours of incident white light from the Sun to the same extent and hence when the scattered light reaches our eyes, the clouds are seen white.<\/p>\n\n\n\n<p><strong>11. Give reason why the smoke from a fire looks white.<\/strong><br><strong><br>Answer<\/strong>: The smoke from a fire looks white because the molecules of smoke are bigger than the wavelength of light, so they scatter lights of all colours equally and the scattered light appears white.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Get summaries, questions, answers, solutions, notes, extras, Concise\/Selina workbook solutions, PDF and guide of chapter 6 Spectrum: ICSE Class 10 Physics which is part of the syllabus of students studying under the Council for the Indian School Certificate Examinations board. These solutions, however, should only be treated as references and can be modified\/changed.&nbsp; Summary When&hellip; <a class=\"more-link\" href=\"https:\/\/mockupbw.site\/2025\/onlinefreenotes\/spectrum-icse-class-10-physics-notes\/\">Continue reading <span class=\"screen-reader-text\">Spectrum: ICSE Class 10 Physics solutions, notes<\/span><\/a><\/p>\n","protected":false},"author":1044,"featured_media":27649,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[49,34],"tags":[28,41,33,35,36,88,37,1209,38,105],"class_list":["post-26024","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-icse","category-notes","tag-answers","tag-hslc","tag-notes","tag-pdf","tag-questions","tag-selina","tag-solutions","tag-spectrum","tag-summary","tag-workbook","entry"],"acf":[],"_links":{"self":[{"href":"https:\/\/mockupbw.site\/2025\/onlinefreenotes\/wp-json\/wp\/v2\/posts\/26024","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/mockupbw.site\/2025\/onlinefreenotes\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/mockupbw.site\/2025\/onlinefreenotes\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/mockupbw.site\/2025\/onlinefreenotes\/wp-json\/wp\/v2\/users\/1044"}],"replies":[{"embeddable":true,"href":"https:\/\/mockupbw.site\/2025\/onlinefreenotes\/wp-json\/wp\/v2\/comments?post=26024"}],"version-history":[{"count":2,"href":"https:\/\/mockupbw.site\/2025\/onlinefreenotes\/wp-json\/wp\/v2\/posts\/26024\/revisions"}],"predecessor-version":[{"id":30027,"href":"https:\/\/mockupbw.site\/2025\/onlinefreenotes\/wp-json\/wp\/v2\/posts\/26024\/revisions\/30027"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/mockupbw.site\/2025\/onlinefreenotes\/wp-json\/wp\/v2\/media\/27649"}],"wp:attachment":[{"href":"https:\/\/mockupbw.site\/2025\/onlinefreenotes\/wp-json\/wp\/v2\/media?parent=26024"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/mockupbw.site\/2025\/onlinefreenotes\/wp-json\/wp\/v2\/categories?post=26024"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/mockupbw.site\/2025\/onlinefreenotes\/wp-json\/wp\/v2\/tags?post=26024"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}